Homework Set #1 Solutions Chapter 15
15.4
The attractive forces exerted on the positive charge by
the negative charges are shown in the sketch and have
magnitudes
F1
and
ke q 2
a2
F2
15.6
ke q 2
F3
2
a 2
ke q 2
2a 2
Fx
F2
F3 cos 45
ke q 2
a2
ke q 2
0.707
2a 2
1.35
ke q 2
a2
Fy
F1
F3 sin 45
ke q 2
a2
ke q 2
0.707
2a 2
1.35
ke q 2
a2
ke q 2
and
a2
tan
FR
so
and
FR
Fx
2
Fy
ke q 2
a2
1.91
2
1.91
Fy
1
tan
Fx
1
1
45
along the diagonal toward the negative charge
The attractive force between the charged ends tends to compress the molecule. Its
magnitude is
F
ke 1e
2
r2
N m2
8.99 10
C2
9
1.60 10
2.17 10
19
6
C
m
2
2
4.89 10
The compression of the “spring” is
x
0.010 0 r
0.010 0 2.17 10 6 m
so the spring constant is k
F
x
4.89 10 17 N
2.17 10 8 m
2.17 10 8 m ,
2.25 10
9
N m
17
N.
15.10
The forces are as shown in the sketch
at the right.
F1
F2
F3
ke q1q2
r122
6.00 10
N m2
8.99 10
C2
9
ke q1 q3
N m2
8.99 10
C2
6.00 10
r
ke q2 q3
N m2
C2
1.50 10
8.99 109
2
23
r
The net force on the 6 C charge is F6
15.12
6
6
C 2.00 10
89.9 N
6
C
43.2 N
2
C 2.00 10
2.00 10-2 m
F2
C
2
5.00 10-2 m
F1
6
C 1.50 10
3.00 10-2 m
9
2
13
6
6
C
67.4 N
2
46.7 N (to the left)
The net force on the 1.5 C charge is F1.5
F1
F3
157 N (to the right)
The net force on the 2 C charge is F 2
F2
F3
111 N (to the left)
Consider the arrangement of charges shown in the sketch at
the right. The distance r is
r
2
0.500 m
0.500 m
2
0.707 m
The forces exerted on the 6.00 nC charge are
F2
8.99 109
2.16 10
and
F3
7
8.99 109
N m2
C2
6.00 10
9
C 2.00 10
0.707 m
9
C
9
C
2
N
N m2
C2
6.00 10
9
C 3.00 10
0.707 m
2
3.24 10
7
N
Thus,
Fx
F2
F3 cos 45.0
and
Fy
F2
F3 sin 45.0
3.81 10
7
N
7.63 10 8 N
The resultant force on the 6.00 nC charge is then
or
15.20
2
FR
Fx
Fy
FR
3.89 10
7
2
3.89 10
N at
tan
1
Fy
Fx
11.3
N at 11.3 below +x axis
The force an electric field exerts on a positive change is in the direction of the field. Since
this force must serve as a retarding force and bring the proton to rest, the force and
hence the field must be in the direction opposite to the proton’s velocity .
The work-energy theorem, Wnet
qE
15.21
7
x
0 KEi
1.60 10
or
19
a
F
m
qE
mp
(b) t
v
a
1.20 106 m s
6.12 1010 m s2
(a)
(c)
x
(d) KE f
v 2f
v02
2a
1
m p v 2f
2
KEi , gives the magnitude of the field as
KE f
KEi
q x
E
C 640 N C
1.673 10-27 kg
1.96 10 5 s
1.20 106 m s
10
2 6.12 10
1
1.673 10
2
2
0
ms
27
2
3.25 10 15 J
1.60 10-19 C 1.25 m
1.63 104 N C
6.12 1010 m s2
19.6 s
11.8 m
kg 1.20 106 m s
2
1.20 10
15
J
15.23
If the resultant field is zero, the
contributions from the two charges must
be in opposite directions and also have
equal magnitudes. Choose the line
connecting the charges as the x-axis, with
the origin at the –2.5 C charge. Then, the
two contributions will have opposite
directions only in the regions x 0 and
x 1.0 m . For the magnitudes to be equal, the point must be nearer the smaller charge.
Thus, the point of zero resultant field is on the x-axis at x 0 .
Requiring equal magnitudes gives
Thus,
2.5
6.0
1.0 m d
ke q1
r12
ke q2
2.5 C
or
d2
r22
6.0 C
1.0 m d
d
Solving for d yields
d
15.36
(a)
F
1.8 m ,
ke q1 q2
r2
or
1.8 m to the left of the
ke e 2
r2
8.99 109 N m 2 C 2 1.60 10
0.53 10
(b) F
v
meac
r F
me
2.5 C charge
10
m
19
C
2
2
8.2 10
8
N
me v 2 r , so
0.53 10
10
m 8.2 10
9.11 10
-31
kg
8
N
2.2 106 m s
2
15.38
Consider the free-body diagram shown at the right.
Fy
0
T cos
Fx
0
Fe
Since Fe
T sin
mg tan
qE , we have
qE
mg tan , or q
mg tan
E
2.00 10 3 kg 9.80 m s2 tan15.0
q
15.45
mg
cos
mg or T
5.25 10 6 C
1.00 103 N C
5.25 C
The sketch at the right gives a free-body diagram of the
positively charged sphere. Here, F1
ke q
2
r 2 is the attractive
force exerted by the negatively charged sphere and F2
is exerted by the electric field.
mg
Fy 0
T cos10 mg or T
cos10
Fx
0
F2
F1 +T sin10
ke q
r2
or qE
qE
2
mg tan10
At equilibrium, the distance between the two spheres is r
E
ke q
4 L sin10
2
2 L sin10 . Thus,
mg tan10
q
8.99 109 N m2 C2 5.0 10
4 0.100 m sin10
8
2.0 10 3 kg 9.80 m s2 tan10
C
2
or the needed electric field strength is
5.0 10
E
4.4 105 N C
8
C